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Dayo Ogundipe edited begin_itemize_item_textbf_Show__.tex
over 8 years ago
Commit id: 80e8ca91b1efdd82bce2493d2b7d5151f74db8bf
deletions | additions
diff --git a/begin_itemize_item_textbf_Show__.tex b/begin_itemize_item_textbf_Show__.tex
index 0e7ba19..2bdb049 100644
--- a/begin_itemize_item_textbf_Show__.tex
+++ b/begin_itemize_item_textbf_Show__.tex
...
\item \textbf{Show that $\log_{10}(2)$ is not a rational number.}
\[ \log_{10}(2) = \frac{p}{q}\] where p,q are integers
\[ 10^{\frac{p}{q}} = 2\]
\[
10^{p} - 10^{q} 10^{p-q}} = 2 \]
$ \nexists$ two integers for p,q.
\\Also
\[ (10^{\frac{p}{q}})^q = 2^q\]
...
$m = a_{1}^{\mu_{1}} , a_{2}^{\mu_{2}}, a_{3}^{\mu_{3}}, a_{i}^{\mu_{i}}$ where $\mu_{i}$ are integers $\ge 1$ and $a_{i}$ are primes.
\\ $n = b_{1}^{v_{1}} , b_{2}^{v_{2}}, b_{3}^{v_{3}}, b_{i}^{v_{i}}$ where $v_{i}$ are integers $\ge 1$ and $b_{i}$ are primes.
\[
m^p - m^q\] m^{p-q} = n\]
\end{itemize}